Ch4. Hilbert Space: An Introduction (Ex.29 ~ Ex.35)

(a) Show that the range of \(\lambda I-T\) defined by \(\{g\in\mathcal{H}: g=(\lambda I-T)f\ \text{for some }f\in\mathcal{H}\}\) is closed.

proof
① \(\lambda I-T\) is continuous, so bounded

Consider \(\{f_n\}\subset\mathcal{H}\) such that \(\lim_{n\to\infty}\|f_n-f\|_{\mathcal{H}}=0\) for \(f\in\mathcal{H}\).

\[\begin{aligned} \|(\lambda I-T)(f_n)-(\lambda I-T)(f)\|_{\mathcal{H}} &=\|\lambda(f_n-f)-T(f_n-f)\|_{\mathcal{H}} \\ &\le \|\lambda(f_n-f)\|_{\mathcal{H}}+\|T(f_n-f)\|_{\mathcal{H}}\qquad(\because\ \Delta\ \text{ineq}) \\ &=|\lambda|\,\|f_n-f\|_{\mathcal{H}}+\|T(f_n-f)\|_{\mathcal{H}} \\ &\le |\lambda|\,\|f_n-f\|_{\mathcal{H}}+\|T\|\,\|f_n-f\|_{\mathcal{H}}\qquad(T\ \text{is bounded}). \end{aligned}\]

Then, \(\lim_{n\to\infty}\|(\lambda I-T)(f_n)-(\lambda I-T)(f)\|_{\mathcal{H}}=0\).

This holds for all \(\{f_n\}\) with \(\|f_n-f\|_{\mathcal{H}}\to 0\), so \(\lambda I-T\) is continuous \(\Rightarrow\) bounded.

② \(V_\lambda\) is closed, where \(V_\lambda\) is kernel of \(\lambda I-T\)

Consider \(\{f_n\}\subset V_\lambda\) with \(\|f_n-f\|_{\mathcal{H}}\to 0\) for \(f\in\mathcal{H}\).

\((\lambda I-T)f_n=0\), so \(\|(\lambda I-T)f\|_{\mathcal{H}}\le |\lambda|\|f_n-f\|_{\mathcal{H}}+\|T\|\|f_n-f\|_{\mathcal{H}}\) (using ①).

\(\|f_n-f\|_{\mathcal{H}}\to 0\) implies \(\|(\lambda I-T)f\|_{\mathcal{H}}=0\), so \((\lambda I-T)f=0\) & \(f\in V_\lambda\).

Thus, \(V_\lambda\) is a closed subspace. Note that \(\mathcal{H}=V_\lambda\oplus V_\lambda^\perp\).

③ \(S:V_\lambda^\perp\to R(\lambda I-T)\), then \(\exists\delta>0\) s.t. \(\|Sx\|_{\mathcal{H}}\ge \delta\|x\|_{\mathcal{H}}\ \forall x\in V_\lambda^\perp\ (x\ne 0)\)

Let \(S:V_\lambda^\perp\to R(\lambda I-T)=:R\) be an operator ‘\(\lambda I-T\)’ between \(V_\lambda\) and \(R\). Assume that for every \(\delta>0\) \(\exists x\in V_\lambda^\perp\) s.t. \(\|Sx\|_{\mathcal{H}}<\delta\|x\|_{\mathcal{H}}\) & \(x\ne 0\).

1. \(\|S(x_n)\|_{\mathcal{H}}\to 0\) Let \(\{x_n\}\) be a bounded sequence in \(V_\lambda^\perp\) s.t. \(\|x_n\|_{\mathcal{H}}=1\), \(\|Sx_n\|_{\mathcal{H}}<\frac{1}{n}\|x_n\|_{\mathcal{H}}\ \forall n\in\mathbb{N}\) \((\delta=\frac{1}{n})\). This means \(\lim_{n\to\infty}\|Sx_n\|=0\). cf) \(S=\lambda I-T\) for \(V_\lambda^\perp\), so \(\lim_{n\to\infty}\|S(x_n)-0\|_{\mathcal{H}}=0\) (∵①, continuous & \(S(0)=0\)).

2. \(T(x_{n_k})\to y\) & \(S(x_{n_k})+T(x_{n_k})=\lambda x_{n_k}\) \(\{x_n\}\) is bounded, and \(T\) is compact, so there is \(\{x_{n_k}\}\) such that \(T(x_{n_k})\) converges: \(\lim_{k\to\infty}\|T(x_{n_k})-y\|_{\mathcal{H}}=0\) for some \(y\in\mathcal{H}\). Then, \(S(x_{n_k})+T(x_{n_k})=\lambda x_{n_k}-T(x_{n_k})+T(x_{n_k})=\lambda x_{n_k}\).

3. \(\dfrac{y}{\lambda}\in V_\lambda^\perp\) : \(\|\lambda x_{n_k}-y\|_{\mathcal{H}}\to 0\) \[\begin{aligned} \|\lambda x_{n_k}-y\|_{\mathcal{H}} &=\|S(x_{n_k})+T(x_{n_k})-y\|_{\mathcal{H}} \\ &\le \|S(x_{n_k})-0\|_{\mathcal{H}}+\|T(x_{n_k})-y\|_{\mathcal{H}}, \end{aligned}\] so \(\lim_{k\to\infty}\|\lambda x_{n_k}-y\|_{\mathcal{H}}=0\) & \(\lim_{k\to\infty}\|x_{n_k}-y/\lambda\|_{\mathcal{H}}=0\). \(\{x_{n_k}\}\subset V_\lambda^\perp\) and \(\mathcal{H}\) is closed implies \(y/\lambda\in V_\lambda^\perp\).

4. \(\dfrac{y}{\lambda}\in V_\lambda\) \[\|S(y/\lambda)-0\|_{\mathcal{H}}\le \|S(y/\lambda)-S(x_{n_k})\|_{\mathcal{H}}+\|S(x_{n_k})-0\|_{\mathcal{H}}.\] \(S\) is continuous, so \(\lim_{k\to\infty}\|S(x_{n_k})-S(y/\lambda)\|_{\mathcal{H}}=0\). Since \(\|S(x_{n_k})\|_{\mathcal{H}}\to 0\), \(\|S(y/\lambda)\|_{\mathcal{H}}=0\) and \(S(y/\lambda)=0\). This means \(y/\lambda\in V_\lambda\).

5. Then, \(y/\lambda\in V_\lambda\cap V_\lambda^\perp\), so \(y=0\)

6. \(y\ne 0\), contradiction \[\|y\|_{\mathcal{H}}\le \|y-\lambda x_{n_k}\|_{\mathcal{H}}+\|\lambda x_{n_k}\|_{\mathcal{H}}=\|y-\lambda x_{n_k}\|_{\mathcal{H}}+|\lambda|,\] and \[|\lambda|=\|\lambda x_{n_k}\|_{\mathcal{H}}\le \|\lambda x_{n_k}-y\|_{\mathcal{H}}+\|y\|_{\mathcal{H}}.\] This means \(\|y\|_{\mathcal{H}}=|\lambda|\), so \(y\ne 0\), which is a contradiction.

④ \(S\) is bijection

1. injective (1-1) Consider \(g_1,g_2\in R(\lambda I-T)\) s.t. \(g_1=g_2\). \[(\lambda I-T)f_1=g_1=g_2=(\lambda I-T)f_2\qquad(f_1,f_2\in\mathcal{H}).\] \(f_1=f_1^\lambda+f_1^\perp\) and \(f_2=f_2^\lambda+f_2^\perp\) \((f_1^\lambda,f_2^\lambda\in V_\lambda\ \&\ f_1^\perp,f_2^\perp\in V_\lambda^\perp)\), so \(g_1=(\lambda I-T)f_1^\perp\) and \(g_2=(\lambda I-T)f_2^\perp\). This means \(T(f_1^\perp-f_2^\perp)=\lambda(f_1^\perp-f_2^\perp)\), so \(f_1^\perp-f_2^\perp\in V_\lambda\). Since \(V_\lambda^\perp\) is a subspace, \(f_1^\perp-f_2^\perp\in V_\lambda^\perp\). Using the fact \(V_\lambda\cap V_\lambda^\perp=\{0\}\), \(f_1^\perp-f_2^\perp=0\). Hence, \(S\) is injective (1-1).

2. surjective (onto) Consider \(g\in R\): \(g=(\lambda I-T)f\) for some \(f\in\mathcal{H}\). \(f=f_\lambda+f_\perp\) where \(f_\lambda\in V_\lambda\) and \(f_\perp\in V_\lambda^\perp\) (∵ \(\mathcal{H}=V_\lambda\oplus V_\lambda^\perp\)), so \(g=(\lambda I-T)(f_\lambda+f_\perp)=(\lambda I-T)f_\perp\). In summary, for \(g\in R\), \(\exists f_\perp\in V_\lambda^\perp\) s.t. \(g=S(f_\perp)\). Thus, \(S\) is surjective (onto). ∴ \(S\) is bijection.

⑤ Convergence in norm implies Cauchy sequence in norm

Let \(\{h_n\}\subset\mathcal{H}\) s.t. \(\lim_{n\to\infty}\|h_n-h\|_{\mathcal{H}}=0\) for \(h\in\mathcal{H}\). For given \(\varepsilon>0\) there is \(N_\varepsilon\in\mathbb{N}\) such that \(n\ge N_\varepsilon\) implies \(\|h_n-h\|_{\mathcal{H}}<\varepsilon\).

\(\|h_m-h_n\|_{\mathcal{H}}\le \|h_m-h\|_{\mathcal{H}}+\|h-h_n\|_{\mathcal{H}}\), so \(\|h_m-h_n\|_{\mathcal{H}}<2\varepsilon\) for \(m,n\ge N_\varepsilon\). So, \(\{h_n\}\) is a Cauchy sequence in \(\mathcal{H}\).

⑥ \(R\) is closed

1. \(\{g_n\}\subset R\) s.t. \(\|g_n-g\|_{\mathcal{H}}\to 0\) Consider \(\{g_n\}\subset R\) such that \(\lim_{n\to\infty}\|g_n-g\|_{\mathcal{H}}=0\) for \(g\in\mathcal{H}\). Choose \(f_n\in V_\lambda^\perp\) such that \(g_n=(\lambda I-T)f_n\). (S is bijective ④) \(\{g_n\}\subset\mathcal{H}\), so it’s a Cauchy sequence (∵⑤). Then, for given \(\varepsilon>0\) there is \(N_\varepsilon\in\mathbb{N}\) such that \(m,n\ge N_\varepsilon\) implies \(\|g_m-g_n\|_{\mathcal{H}}<\varepsilon\).

2. \(\{f_n\}\) is Cauchy in \(V_\lambda^\perp\) Using ③, there is \(\delta>0\) s.t. \(\|S(f_n)\|_{\mathcal{H}}=\|g_n\|_{\mathcal{H}}\ge \delta\|f_n\|_{\mathcal{H}}\). \(f_m-f_n\in V_\lambda^\perp\) (∵ subspace) and \(S(f_m-f_n)=g_m-g_n\) (∵ \(S\) is linear), so \(\|f_m-f_n\|_{\mathcal{H}}\le \frac{1}{\delta}\|g_m-g_n\|_{\mathcal{H}}\). Then, \(m,n\ge N_\varepsilon\) implies \(\|f_m-f_n\|_{\mathcal{H}}<\frac{1}{\delta}\varepsilon\). (∵⑥-1), Cauchy) This holds for all \(\varepsilon>0\), so \(\{f_n\}\) is a Cauchy sequence in \(V_\lambda^\perp\).

3. \(g\in R\), conclusion \(V_\lambda^\perp\) is a closed subspace, so it is a Hilbert space which is complete. Then, there is \(f\in V_\lambda^\perp\) such that \(\lim_{n\to\infty}\|f_n-f\|_{\mathcal{H}}=0\). \(S\) is continuous, so \(\lim_{n\to\infty}\|S(f_n)-S(f)\|_{\mathcal{H}}=\lim_{n\to\infty}\|g_n-S(f)\|_{\mathcal{H}}=0\). This means \(S(f)=g\), thus \(g\in R\).

∴ \(R\) is closed.

(b) Show, by example, that this may fail when \(\lambda=0\)

proof
It suffices to show the existence of operator \(T\) that is compact but \(R(T)\) is not closed. Let \(\{e_n\}\) be an orthonormal basis for \(H\) and let \[ Tx=\sum_{n=1}^\infty \frac{1}{n}(x,e_n)e_n \qquad (x\in H). \]

① \(T\) is linear \((f,g\in H,\ \alpha,\beta\in\mathbb{C})\) \[ \begin{aligned} T(\alpha f+\beta g) &=\sum_{n=1}^\infty \frac{1}{n}(\alpha f+\beta g,e_n)e_n \\ &=\sum_{n=1}^\infty \frac{1}{n}\{\alpha(f,e_n)e_n+\beta(g,e_n)e_n\} \\ &=\alpha\sum_{n=1}^\infty \frac{1}{n}(f,e_n)e_n+\beta\sum_{n=1}^\infty \frac{1}{n}(g,e_n)e_n \\ &=\alpha T(f)+\beta T(g). \end{aligned} \]

② \(T\) is compact

1. Choose \(\{z_k\}\subset H\) s.t. \((z_k,e_n)\to a_n\)
   Let \(\{x_m\}=X_0\subset H\) be a bounded sequence s.t. \(\|x_m\|_H\le 1\) \(\forall m\in\mathbb{N}\). Then \(|(x_m,e_1)|\le \|x_m\|\|e_1\|\le 1\). This means \(\{(x_m,e_1)\}_m\) is a bounded sequence, so there is a subsequence \(\{x_{1,k}\}_k=X_1\subset X_0\) and \(a_1\in\mathbb{C}\) s.t. \(\lim_{k\to\infty}(x_{1,k},e_1)=a_1\) \((|a_1|\le 1)\). Then, for every fixed \(n\in\mathbb{N}\) there exists \(a_n\in\mathbb{C}\) and subsequence \(\{x_{n,k}\}_k=X_n\) s.t. \(\lim_{k\to\infty}(x_{n,k},e_n)=a_n\) \((X_n\subset X_{n-1},\ |a_n|\le 1)\). Let \(z_k=x_{k,k}\) for simplicity, then \(\lim_{k\to\infty}(z_k,e_n)=a_n\) \((n\in\mathbb{N})\). \(\{z_k\}\subset\{x_m\}_m\).

2. \(\left\|Tz_k-\sum_{n=1}^\infty \frac{1}{n}a_ne_n\right\|_H\to 0\), conclusion \[\begin{aligned} T(z_k)-\sum_{n=1}^\infty \frac{1}{n}a_ne_n &=\sum_{n=1}^\infty \left\{\frac{1}{n}(z_k,e_n)e_n-\frac{1}{n}a_ne_n\right\} \\ &=\sum_{n=1}^\infty \frac{1}{n}\big((z_k,e_n)-a_n\big)e_n. \end{aligned} \] Then, \[ \begin{aligned} \left\|Tz_k-\sum_{n=1}^\infty \frac{1}{n}a_ne_n\right\|_H^2 &=\sum_{n=1}^\infty \left|\frac{1}{n}\big((z_k,e_n)-a_n\big)\right|^2 \\ &=\sum_{n=1}^\infty \frac{1}{n^2}\big|(z_k,e_n)-a_n\big|^2, \end{aligned} \] and \[ \begin{aligned} \lim_{k\to\infty}\left\|Tz_k-\sum_{n=1}^\infty \frac{1}{n}a_ne_n\right\|_H^2 &=\lim_{k\to\infty}\sum_{n=1}^\infty \frac{1}{n^2}\big|(z_k,e_n)-a_n\big|^2 \\ &=\sum_{n=1}^\infty \lim_{k\to\infty}\frac{1}{n^2}\big|(z_k,e_n)-a_n\big|^2 \quad (\mathrm{DCT})\\ &=0. \end{aligned} \] This shows \(\left\|Tz_k-\sum_{n=1}^\infty \frac{1}{n}a_ne_n\right\|_H\to 0\) and this holds for all bounded \(\{x_m\}\), so \(T\) is compact.

cf) \(\left\|\sum_{n=1}^\infty \frac{1}{n}a_ne_n\right\|^2=\sum_{n=1}^\infty \frac{1}{n^2}|a_n|^2\le \sum_{n=1}^\infty \frac{1}{n^2}<\infty\), so \(\sum_{n=1}^\infty \frac{1}{n}a_ne_n\in H\).

③ \(R(T) := R\) is not closed

1. \(\{T(\psi_m)\}_m\subset R\) converges
   Consider \(\psi_m=\sum_{k=1}^m e_k\), and \(y=\sum_{n=1}^\infty \frac{1}{n}e_n\). \[ \begin{aligned} T(\psi_m) &=\sum_{n=1}^\infty \frac{1}{n}(\psi_m,e_n)e_n =\sum_{n=1}^\infty \frac{1}{n}\sum_{k=1}^m (e_k,e_n)e_n =\sum_{n=1}^m \frac{1}{n}e_n. \end{aligned} \] Then, \[ \|T(\psi_m)-y\|_H^2 =\left\|\sum_{n=m+1}^\infty \frac{1}{n}e_n\right\|_H^2 =\sum_{n=m+1}^\infty \frac{1}{n^2} =\sum_{n=1}^\infty \frac{1}{n^2}-\sum_{n=1}^m \frac{1}{n^2}. \] Thus, \(\lim_{m\to\infty}\|T(\psi_m)-y\|_H=0\) \(\left(\sum \frac{1}{n^2}<\infty\right)\).

2. \(y\notin R\)
   Assume that \(y\in R\), then there is \(z\in H\) such that \(Tz=y\): \[ \sum_{n=1}^\infty \frac{1}{n}(z,e_n)e_n=\sum_{n=1}^\infty \frac{1}{n}e_n. \] \((z,e_n)=1\) for all \(n\in\mathbb{N}\), so \(z=\sum_{n=1}^\infty e_n\). Then, \[ \|z\|_H^2=\left\|\sum_{n=1}^\infty e_n\right\|_H^2=\sum_{n=1}^\infty \|e_n\|_H^2=\infty, \] contradiction. \((z\notin H)\)

\(\therefore\ R\) is not closed.

(c) Show that the range of \(\lambda I-T\) is \(H\) if and only if the null space of \(\overline{\lambda}I-T^*\) is trivial

proof
① \(\Leftarrow\)

Assume that \(R\ne H\), then \(R^\perp\ne\{0\}\). \(H=R\oplus R^\perp\), so choose \(y\in R^\perp\) s.t. \(y\ne 0\). Then \((y,(\lambda I-T)z)=0\) for all \(z\in H\) \((\because (\lambda I-T)z\in R)\).
\[ ((\lambda I-T)^*y,z)=((\overline{\lambda}I-T^*)y,z)=0 \quad (\text{adjoint}) \qquad \forall z\in H. \] Applying \((\overline{\lambda}I-T^*)y\) to \(z\), \(\|(\overline{\lambda}I-T^*)y\|_H=0\) so \((\overline{\lambda}I-T^*)y=0\).
Thus, there is \(y\ne 0\) s.t. \(\overline{\lambda}y=T^*y\) (non-trivial solution, contradiction).

② \(\Rightarrow\)

\(R=H\) and consider \(y\in H\) with \(\overline{\lambda}y=T^*y\) \((y:\) null space of \(\overline{\lambda}I-T^*)\). Then \(((\overline{\lambda}I-T^*)y,z)=0\) for all \(z\in H\), as well as \((y,(\lambda I-T)z)=0\).
\((\lambda I-T)z\in R=H\), so \(y\in R^\perp\). Since \(y\in R\) & \(R\cap R^\perp=\{0\}\), \(y=0\).
Thus, the null space of \(\overline{\lambda}I-T^*\) is trivial.

cf) \((aT)^*=\overline{a}T^*\) \[ \langle aTx,y\rangle=\langle Tx,\overline{a}y\rangle=\langle x,T^*(\overline{a}y)\rangle=\langle x,\overline{a}T^*y\rangle. \]

(a) Show that \(T\) is a bounded operator on \(H:\ \|T\|=\sup_n|\lambda_n|\)

proof
① def of \(S_N(f),\,S_N(Tf)\)

Let \(S_N(f)=\sum_{n=-N}^{N} a_n e^{inx}\) and \(S_N(Tf)=\sum_{n=-N}^{N}\lambda_n a_n e^{inx}.\)
It is known that \(\lim_{N\to\infty}\|S_N(f)-f\|_2=0\) and \(\lim_{N\to\infty}\|S_N(Tf)-Tf\|_2=0\) (thm 4.3.2 iii)

② \(S_N(e^{inx}),\ S_N(Te^{inx})\)

When \(f(x)=e^{inx}\), Fourier coefficient is \(a_n=\frac{1}{2\pi}\int_{-\pi}^{\pi} e^{inx}e^{-inx}\,dx=1\) and \(a_m=0\) \((m\neq n).\) Since \(f\sim e^{inx}\), \(Tf\sim \lambda_n e^{inx}.\)
Then, \(S_N(e^{inx})=e^{inx}\) and \(S_N(Te^{inx})=\lambda_n e^{inx}\) \((N\ge |n|)\) by def of \(S_N\).

③ \(Te^{inx}=\lambda_n e^{inx}\)

\[\begin{aligned} \|T(e^{inx})-\lambda_n e^{inx}\|_2 &\le \|T(e^{inx})-S_N(T(e^{inx}))\|_2+\|S_N(T(e^{inx}))-\lambda_n e^{inx}\|_2\\ &=\|T(e^{inx})-S_N(T(e^{inx}))\|_2\quad (N\ge |n|,\ \because ②) \end{aligned}\]

Letting \(N\to\infty\), \(\|T(e^{inx})-\lambda_n e^{inx}\|_2=0\) \((n\in\mathbb{Z},\ \because ①)\)
This means \(T(e^{inx})=\lambda_n e^{inx}\) for all \(n\in\mathbb{Z}\), so \(\lambda_n\) is an eigenvalue with on eigenvector \(e^{inx}\).

④ \(T\) is bounded.

Thus, \(\|T\|=\sup_n|\lambda_n|\) which is finite (\(\lambda_n\) is bounded), and \(T\) is bounded.

(b) Let \(\tau_h(f(x))=f(x-h)\) \((h\in\mathbb{R})\). Show that \(T\circ\tau_h=\tau_h\circ T\)

proof
① \(\tau_h\) is linear and bounded

\(\tau_h(\alpha f(x)+\beta g(x))=\alpha f(x-h)+\beta g(x-h)=\alpha\tau_h(f)+\beta\tau_h(g)\), so \(\tau_h\) is linear.
\(\|\tau_h(f)\|_2=\|f(x-h)\|_2=\|f\|_2\), so \(\|\tau_h\|=1\ \to\) bounded (& continuous)

② \(S_N(Tf(x-h))=\tau_h(S_N(Tf(x)))\)

Fourier coefficient is translation-invariant, so \(f(x-h)\sim \sum_{n=-\infty}^{\infty} a_n e^{in(x-h)}\) and \(Tf(x-h)\sim \sum_{n=-\infty}^{\infty}\lambda_n a_n e^{in(x-h)}.\)
Then, \(S_N(Tf(x-h))=\sum_{n=-N}^{N}\lambda_n a_n e^{in(x-h)}\)

\(\tau_h(S_N(Tf(x)))=\tau_h\!\left(\sum_{n=-N}^{N}\lambda_n a_n e^{inx}\right)=\sum_{n=-N}^{N}\lambda_n a_n e^{in(x-h)}\ \ (\because\ \tau_h\ \text{is linear},\ ①)\)

This proves \(S_N(Tf(x-h))=\tau_h(S_N(Tf)).\)

③ \(T\circ\tau_h=\tau_h\circ T\), conclusion

\[\begin{aligned} \|T\circ\tau_h(f)-\tau_h\circ T(f)\|_2 &=\|Tf(x-h)-\tau_h(Tf(x))\|_2\\ &\le\underbrace{ \|Tf(x-h)-S_N(Tf(x-h))\|_2}_{1)}+\underbrace{\|\tau_h(S_N(Tf(x)))-\tau_h(Tf(x))\|_2}_{2)} \end{aligned}\]

By def of \(S_N\), \(\lim_{N\to\infty}\|S_N(Tf(x-h))-T\circ\tau_h(f)\|_2=0\ \to 1)\)

\(\lim_{N\to\infty}\|S_N(Tf(x))-Tf(x)\|_2=0\) and \(\tau_h\) is continuous \((\because ①)\), so \(\lim_{N\to\infty}\|\tau_h(S_N(Tf))-\tau_h(Tf)\|_2=0\ \to 2)\)

Letting \(N\to\infty\), \(\|T\circ\tau_h(f)-\tau_h\circ T(f)\|_2=0\)

Thus, \(T\circ\tau_h=\tau_h\circ T\),

(c) \(T\) is a bounded operator on \(H\) that commutes with translations. Show that \(T\) is a Fourier multiplier operator,

proof
① \(T(e^{inx})=b_n e^{inx}\)

Consider \(e^{inx}\) and let \(T(e^{inx})\sim \sum_{k=-\infty}^{\infty}\lambda_k e^{ikx}\)

\(T\circ\tau_h(e^{inx})=\tau_h\circ T(e^{inx})\sim \sum_{k=-\infty}^{\infty}\lambda_k e^{ik(x+h)}\ \ (\because\ \text{translation invariance of F-C})\)
\(T(e^{in(x+h)})=e^{-inh}T(e^{inx})\sim \sum_{k=-\infty}^{\infty}\lambda_k e^{-inh}\cdot e^{ikx}\ \ (\because\ \text{linearity of F-C})\)

Then \(\lambda_k e^{ikh}=\lambda_k e^{-inh}\), so \(\lambda_k=0\ (k\neq n)\)
This implies \(T(e^{inx})\sim \lambda_n e^{inx}\), so \(T(e^{inx})=\lambda_n e^{inx}\ (\because (a)\ ②)\ \forall\, n\in\mathbb{Z}.\)

② conclusion

\(\{e^{inx}\}_n\) is an orthonormal basis for \(H\), so \(f\in H\) can be written as follows: \(f(x)=\sum_{n=-\infty}^{\infty} a_n e^{inx}.\)

\(\frac{1}{2\pi}\int_{-\pi}^{\pi} f(x)e^{-imx}\,dx=\frac{1}{2\pi}\int_{-\pi}^{\pi} a_m e^{imx}e^{-imx}\,dx=a_m\), so \(\{a_n\}\) is Fourier coefficient : \(f(x)\sim \sum_{n=-\infty}^{\infty} a_n e^{inx}.\)

\(Tf(x)=T\!\left(\sum_{n=-\infty}^{\infty} a_n e^{inx}\right)=\sum_{n=-\infty}^{\infty}\lambda_n a_n e^{inx}\), so \(Tf(x)\sim \sum_{n=-\infty}^{\infty}\lambda_n a_n e^{inx}.\)

Thus, \(T\) is Fourier multiplier operator.

(a)-1 Show that \(F(x)=Tf(x)\) is absolutely continuous

proof
\[2\pi T(f)(x)=\int_{-\pi}^{\pi}K(x-y)f(y)\,dy=\int_{-\pi}^{\pi} i(\mathrm{sgn}(x-y)\pi-(x-y))f(y)\,dy.\] So, \[\begin{aligned} -i2\pi\,T(f)(x)&=\int_{-\pi}^{\pi}\big(\pi\,\mathrm{sgn}(x-y)-(x-y)\big)f(y)\,dy \\ &\biggl( \begin{cases} x-y\ge 0 &\to y\le x\\ x-y\le 0 &\to y\ge x\end{cases} \biggr) \\ &=\int_{-\pi}^{x}\big(\pi-(x-y)\big)f(y)\,dy-\int_{x}^{\pi}\big(\pi+(x-y)\big)f(y)\,dy\\ &=(\pi-x)\int_{-\pi}^{x}f(y)\,dy+\int_{-\pi}^{x}y f(y)\,dy-(\pi+x)\int_{x}^{\pi}f(y)\,dy+\int_{x}^{\pi}y f(y)\,dy\\ &=(\pi-x)\int_{-\pi}^{x}f(y)\,dy-(\pi+x)\int_{x}^{\pi}f(y)\,dy+\int_{-\pi}^{\pi}y f(y)\,dy\\ &=\pi\left\{\int_{-\pi}^{x}f(y)\,dy-\int_{x}^{\pi}f(y)\,dy\right\}-x\left\{\int_{-\pi}^{x}f(y)\,dy+\int_{x}^{\pi}f(y)\,dy\right\}+\int_{-\pi}^{\pi}y f(y)\,dy\\ &=\pi\left\{\int_{-\pi}^{x}f(y)\,dy+\int_{\pi}^{x}f(y)\,dy\right\}-x\left\{\int_{-\pi}^{\pi}f(y)\,dy\right\}+\int_{-\pi}^{\pi}y f(y)\,dy \end{aligned}\]

Since\[\left|\int_{-\pi}^{\pi}y f(y)\,dy\right|\le \int_{-\pi}^{\pi}|y f(y)|\,dy\le \pi\int_{-\pi}^{\pi}|f(y)|\,dy<\infty,\] \(\int_{-\pi}^{\pi}y f(y)\,dy\) is finite. It is obvious that \(\int_{-\pi}^{\pi} f(y)\,dy\) is finite \((\because f\in L^1([-\pi,\pi])).\)

Also, \(\int_{-\pi}^{x}f(y)\,dy\) and \(\int_{\pi}^{x}f(y)\,dy\) are absolutely continuous with respect to \(x\in[-\pi,\pi]\ (\because f\in L^1([-\pi,\pi]))\), and \(x\) is absolutely continuous.

Thus, \(F(x)\) is absolutely continuous on \([-\pi,\pi]\)

(a)-2 If \(\int_{-\pi}^{\pi}f(y)\,dy=0\), show that \(F'(x)=if(x)\) a.e \(x\)

proof
\[(-i2\pi F(x))'=\pi\{f(x)+f(x)\}-\int_{-\pi}^{\pi}f(y)\,dy=2\pi f(x)-\int_{-\pi}^{\pi}f(y)\,dy\quad \text{a.e }x.\] \[\Rightarrow\ F'(x)=\frac{-2\pi}{-i2\pi}f(x)+\frac{1}{i2\pi}\int_{-\pi}^{\pi}f(y)\,dy=if(x)+\frac{1}{i2\pi}\int_{-\pi}^{\pi}f(y)\,dy.\] Since \(\int_{-\pi}^{\pi}f(y)\,dy=0\), \(F'(x)=if(x)\) a.e \(x\).

(b)-1 Show that \(T\) is compact

proof
Let \(K(x,y)=K(x-y)\)

\[\begin{aligned} \int_{-\pi}^{\pi}|K(x-y)|^2\,dx &=\int_{-\pi}^{\pi}\big|\pi\,\mathrm{sgn}(x-y)-(x-y)\big|^2\,dx \quad\left\{\begin{aligned} x-y\ge 0 &: x\ge y\\ x-y\le 0 &: x\le y \end{aligned}\right.\\ &=\int_{-\pi}^{y}|x-y+\pi|^2\,dx+\int_{y}^{\pi}|x-y-\pi|^2\,dx\qquad (x-y:=t)\\ &=\int_{-\pi-y}^{0}(t+\pi)^2\,dt+\int_{0}^{\pi-y}(t-\pi)^2\,dt\\ &=\int_{-\pi-y}^{0}(t^2+2\pi t+\pi^2)\,dt+\int_{0}^{\pi-y}(t^2-2\pi t+\pi^2)\,dt\\ &=\left[\frac{1}{3}t^3+\pi t^2+\pi^2 t\right]_{t=-\pi-y}^{0}+\left[\frac{1}{3}t^3-\pi t^2+\pi^2 t\right]_{t=0}^{t=\pi-y}\\ &=-\left\{\frac{1}{3}(-\pi-y)^3+\pi(-\pi-y)^2+\pi^2(-\pi-y)\right\}+\left\{\frac{1}{3}(\pi-y)^3-\pi(\pi-y)^2+\pi^2(\pi-y)\right\}\\ &=\left\{\frac{1}{3}(\pi+y)^3-\pi(\pi+y)^2+\pi^2(\pi+y)\right\}+\left\{\frac{1}{3}(\pi-y)^3-\pi(\pi-y)^2+\pi^2(\pi-y)\right\}\\ &=\frac{1}{3}(2\pi^3+2\cdot 3\cdot \pi y^2)-\pi(2\pi^2+2y^2)+\pi^2\cdot 2\pi\\ &=\frac{2}{3}\pi^3+2\pi y^2-2\pi^3-2\pi y^2+2\pi^3=\frac{2}{3}\pi^3 \end{aligned}\]

So, \[\int_{-\pi}^{\pi}\int_{-\pi}^{\pi}|K(x-y)|^2\,dx\,dy=\int_{-\pi}^{\pi}\frac{2}{3}\pi^3\,dy=\frac{2}{3}\pi^3\cdot 2\pi=\frac{4}{3}\pi^4<\infty,\] and \(K(x-y)\in L^2([-\pi,\pi]^2)\).

Hence, \(T\) is a Hilbert-Schmidt operator on \(L^2([-\pi,\pi])\), and it is compact.

(b)-2 Show that \(T\) is symmetric on \(L^2([-\pi,\pi])\)

proof Due to prop 4.5.5(iii), \(T^*\) has a kernel \(\overline{K(y,x)}=\overline{K(y-x)}\)

\[K(y,x)=K(y-x)=-i(\pi\,\mathrm{sgn}(y-x)-(y-x))=i(\pi\,\mathrm{sgn}(x-y)-(x-y))=K(x,y)\] Then, \[T^*f(x)=\frac{1}{2\pi}\int_{-\pi}^{\pi}\overline{K(y-x)}f(y)\,dy=\frac{1}{2\pi}\int_{-\pi}^{\pi}K(x-y)f(y)\,dy=Tf(x).\]

This holds for all \(f\in L^1([-\pi,\pi])\), so \(T^*=T\).

(c) Show that \(\psi(x)\in L^2([-\pi,\pi])\) is an eigenfunction for \(T\) if and only if \(\psi(x)\) is equal to \(e^{inx}\) with eigenvalue \(\frac{1}{n}\ (n\ne 0)\), or \(\psi=1\) with \(0\).

proof It suffices to show that eigenfunctions are \(\{1,e^{\pm ix},e^{\pm 2ix},\cdots\}\) with eigenvalues \(\{0,\pm 1,\pm \frac{1}{2},\cdots\}\).

① \(\psi(x)=1\ \to\ \lambda=0\)

\[(T1)'=i\cdot 1+\frac{1}{2\pi i}\int_{-\pi}^{\pi}1\,dy=i\cdot 1+\frac{1}{2\pi i}\cdot 2\pi=i\cdot 1-i\cdot 1=0,\] so \(T1=\lambda 1\) and \(1\) is an eigenvector.

Using the result in (a)-1, \[\{-i2\pi\,T(1)(x)\}=\pi\left\{\int_{-\pi}^{x}f(y)\,dy+\int_{\pi}^{x}f(y)\,dy\right\}-x\int_{-\pi}^{\pi}f(y)\,dy+\int_{-\pi}^{\pi}y f(y)\,dy\Big|_{f(y)=1}\] \[=\int_{-\pi}^{\pi}y\,dy=0.\] So, \(T(1)=0\) and \(\lambda=0\).

② other \(\psi\) with \(\lambda\ne 0\)

1. \(\psi(x)=C\cdot e^{\frac{i}{\lambda}x}\) (lemma 6.6.3) Consider other eigenfunction \(\psi\) orthogonal to 1. Then, \((\psi,1)=0\) and \(\int_{-\pi}^{\pi}\psi(x)\,dx=0.\)
   Then, \((T\psi)'=i\psi\). Let \(\lambda\) be the eigenvalue : \(T\psi=\lambda\psi\), then \(i\psi=\lambda\psi'.\)
   The solution of \(\psi'=\frac{i}{\lambda}\psi\ (\because \lambda\ne 0)\) is \(\psi(x)=C\cdot e^{\frac{i}{\lambda}x}\ \ (C\in\mathbb{C},\ C\ne 0)\)

2. \(\psi\) has period \(2\pi\) \(K\) is \(2\pi\)-periodic, so \[Tf(x+2\pi)=\frac{1}{2\pi}\int_{-\pi}^{\pi}K(x+2\pi-y)f(y)\,dy=\frac{1}{2\pi}\int_{-\pi}^{\pi}K(x-y)f(y)\,dy=Tf(x).\] This makes \(Tf(x)\) \(2\pi\)-periodic for all \(f\).
   Then, \(T\psi=\lambda\psi\) makes \(\psi\) \(2\pi\)-periodic as well.

3. \(\lambda=\pm\frac{1}{n}\) \[\psi(\pi)=\psi(-\pi)\ \Leftrightarrow\ C\cdot e^{\frac{i}{\lambda}\pi}=C\cdot e^{\frac{i}{\lambda}(-\pi)}\ \Leftrightarrow\ e^{\frac{i}{\lambda}2\pi}=1\] \[\Leftrightarrow\ \frac{2\pi}{\lambda}=\pm 2n\pi\ (n\in\mathbb{N})\ \Leftrightarrow\ \lambda=\pm\frac{1}{n}\ (n\in\mathbb{N}).\]

4. Conclusion Thus, \(\{C\cdot e^{inx}\}_{n\in\mathbb{Z}}\) is the eigenvectors with eigenvalues \(\pm\frac{1}{n}\ (n\ne 0)\) and \(0\ (n=0)\).

(d) Show that \(\{e^{inx}\}_{n\in\mathbb{Z}}\) is an orthonormal basis for \(L^2([-\pi,\pi])\)

proof
\(T\) is a compact, symmetric operator \((\because (b))\) on a Hilbert space \(L^2([-\pi,\pi])\) (p.162 ex1). So \(\{C\cdot e^{inx}\}_{n\in\mathbb{Z}}\) is a basis for \(L^2([-\pi,\pi])\) (Hmw 4b.2, spectral thm)

Thus, \(\{e^{inx}\}_{n\in\mathbb{Z}}\) is an orthonormal basis for \(L^2([-\pi,\pi])\)

(a)-1 Show that \(T\) is bounded, linear operator with \(T=T^*\)

proof
① \(T\) is bounded
\[\|Tf\|_2^2=\int_0^1 t^2|f(t)|^2dt\le \int_0^1|f(t)|^2dt=\|f\|_2^2,\] so \(\|T\|\le 1\) and \(T\) is bounded

② \(T\) is linear
Let \(h(t)=\alpha\cdot f(t)+\beta\cdot g(t)\ \ (\alpha,\beta\in\mathbb{C})\)
\(Th(t)=t\cdot h(t)=\alpha\cdot t f(t)+\beta\cdot t g(t)=\alpha\cdot Tf(t)+\beta\cdot Tg(t)\). So \(T\) is linear.

③ \(T^*=T\)
\[(Tf,g)=\int_0^1 t f(t)\overline{g(t)}\,dt=\int_0^1 f(t)\cdot \overline{t\cdot g(t)}\,dt=(f,Tg)=(f,T^*g)\ \ (f,g\in L^2([0,1]))\] Since \(T^*\) is unique, \(T^*=T\).

(a)-2 Show that \(T\) is not compact

proof
① \(f_n(t)=\sqrt{n}\chi_{[1-\frac{1}{n},\,1]}(t)\in L^2([0,1])\)
Consider \[\{f_n\}_{n\in\mathbb{N}}=\left\{\sqrt{n}\chi_{[1-\frac{1}{n},\,1]}(t)\right\}_{n=1}^\infty,\] which are all bounded functions. \(\int_0^1|f_n(t)|^2dt=n\cdot\int_{1-\frac{1}{n}}^1 1\,dt=n\cdot\frac{1}{n}=1\), so \(\|f_n\|_2=1\) and \(\{f_n\}\subset L^2([0,1])\).

② assume \(\|Tf_{n_k}-g\|_2\to 0\)
Assume that there exist a subsequence \(\{f_{n_k}\}\) such that \(\lim_{k\to\infty}\|Tf_{n_k}-g\|_2=0\) for some \(g\in L^2([0,1])\). Then, for every \(\varepsilon>0\) there is \(k_\varepsilon\in\mathbb{N}\) such that \(k\ge k_\varepsilon\) implies \(\|Tf_{n_k}-g\|_2<\varepsilon\).

③ \(g\ne 0\), \(\exists a\) s.t \(\int_0^a|g(t)|^2dt>0\)
\[\begin{aligned} \|Tf_n\|_2^2&=\|t\cdot f_n(t)\|_2^2 \\ &=\int_0^1 t^2\cdot n\chi_{[1-\frac{1}{n},\,1]}(t)\,dt =n\int_{1-\frac{1}{n}}^1 t^2\,dt\\ &=n\left[\frac{1}{3}t^3\right]_{t=1-\frac{1}{n}}^{1} =\frac{n}{3}\left\{1-\left(1-\frac{1}{n}\right)^3\right\} \\ &=\frac{n}{3}\left\{1-\left(1-\frac{3}{n}+\frac{3}{n^2}-\frac{1}{n^3}\right)\right\}\\ &=\frac{n}{3}\left(\frac{3}{n}-\frac{3}{n^2}+\frac{1}{n^3}\right) \\ &=1-\frac{3}{n}+\frac{1}{3n^2}. \end{aligned}\] Since \(\lim_{n\to\infty}\|Tf_n\|_2=1\), \(\lim_{k\to\infty}\|Tf_{n_k}\|_2=1\) as well.
This means \(g\ne 0\) (even in a.e), so there is \(a\in(0,1)\) such that \(\varepsilon^2:=\int_0^a|g(t)|^2dt>0\).

④ \(\|g-Tf_{n_k}\|_2\ge \varepsilon\), contradiction

Choose \(k\) sufficiently larger than \(k_\varepsilon\) so that \(a\le 1-\frac{1}{n_k}\). Then \[\begin{aligned} \|g-Tf_{n_k}\|_2^2 &=\int_0^1|g(t)-t\cdot f_{n_k}(t)|^2dt\\ &=\int_0^a|g(t)|^2dt+\int_a^1|g(t)-t\cdot f_{n_k}(t)|^2dt\\ &\ge \int_0^a|g(t)|^2dt=\varepsilon^2. \end{aligned}\] This is a contradiction to \(\|Tf_{n_k}-g\|<\varepsilon\) for all \(k\ge k_\varepsilon\).
\(\therefore\) \(T\) is not compact

(b) Show that \(T\) has no eigenvectors

proof
Let \(\psi\) be an eigenvector of \(T\) corresponding to \(\lambda\).
\(t\cdot \psi(t)=T\psi(t)=\lambda\cdot \psi(t)\) so \((t-\lambda)\psi(t)=0.\)
This holds for all \(t\in[0,1]\), so \(\psi(t)=0\) (a.e).
Then, \(\psi\) can’t be an eigenvector, which is a contradiction.

(1) Show that the operator \(T\) defined by \(T(\psi_k)=\frac{1}{k}\psi_{k+1}\) is compact

proof
① compact \(T_n\)

Define an operator \(T_n\) by \(T_n(\psi_k)=\frac{1}{k}\psi_{k+1}\) if \(1\le k\le n\) and \(T_n(\psi_k)=0\) for \(k>n.\)
\(T_n\) is of finite rank, so it is compact for all \(n\in\mathbb{N}\).

② \(\|T-T_n\|\to 0\), conclusion

For \(x=\sum_{k=1}^\infty x_k\psi_k\in H\ \ (x_k\in\mathbb{C})\),
\[Tx=T\left(\sum_{k=1}^\infty x_k\psi_k\right)=\sum_{k=1}^\infty x_k\cdot\frac{1}{k}\psi_{k+1}\] and \[T_nx=\sum_{k=1}^\infty x_kT_n(\psi_k)=\sum_{k=1}^n x_k\frac{1}{k}\psi_{k+1}.\]
So, \((T-T_n)x=\sum_{k=n+1}^\infty x_k\cdot\frac{1}{k}\psi_{k+1}.\)

Then, \[\|(T-T_n)x\|_H^2=\sum_{k=n+1}^\infty \frac{1}{k^2}x_k^2\le \frac{1}{(n+1)^2}\sum_{k=n+1}^\infty x_k^2\le \frac{1}{(n+1)^2}\|x\|_H^2.\]

This means \(\|T-T_n\|\le \frac{1}{n+1}\), so \(\lim_{n\to\infty}\|T-T_n\|=0.\) Thus, \(T\) is compact.

(2) Show that \(T\) has no eigenvectors

proof
Assume that \(f\in H\) is an eigenvector with \(f=\sum_{n=1}^\infty a_n\psi_n.\)

\[T(f)=\sum_{n=1}^\infty a_nT(\psi_n)=\sum_{n=1}^\infty a_n\cdot\frac{1}{n}\psi_{n+1}\] and \[Tf=\lambda f=\lambda\sum_{n=1}^\infty a_n\psi_n.\]

Then, \[\lambda a_1\psi_1+\lambda a_2\psi_2+\lambda a_3\psi_3+\cdots=0\cdot\psi_1+a_1\cdot\frac{1}{1}\psi_2+a_2\cdot\frac{1}{2}\psi_3+\cdots\]

So, we can obtain \(a_1=0,\ a_2=0,\ \cdots,\ a_n=0,\ \cdots\) inductively (whether \(\lambda=0\) or not).

So, \(f=0\), which is a contradiction.

(a) If \(T_1\) and \(T_2\) are two linear, symmetric, compact operators on \(H\) that commute \((T_1T_2=T_2T_1)\), show that they can be diagonalized simultaneously: there is an orthonormal basis for \(H\) which consists of eigenvectors for both \(T_1\) and \(T_2\).

proof
① common eigenvectors \(\{\phi_n\}\)

\(T_1,\ T_2\) are compact and symmetric, so there exist orthonormal basis \(\{\phi_{1,k}\},\ \{\phi_{2,k}\}\) of \(H\) that consist of eigenvectors of \(T_1\) and \(T_2\) respectively (spectral thm.). Let \(\{\phi_n\}_n=\{\phi_{1,k}\}\cap\{\phi_{2,k}\}\) be common eigenvectors of \(T_1\) and \(T_2\).

② assume \(S=\operatorname{span}\{\phi_n\}\ne H\) (cf. orthonormal basis is dense in \(H\).)

Let \(S\) be the closure of the subspace spanned by all of the \(\{\phi_n\}\). Assume that \(S\ne H\), then \(H=S\oplus S^\perp\), where \(S^\perp\) has some elements other than \(0\).

③ eigenvector of \(T_1\): \(\psi\in S^\perp \Rightarrow\) eigenvector of \(T_2\)

Let \(\psi\) be an eigenvector of \(T_1\) with eigenvalue \(\lambda\) such that \(\psi\in S^\perp,\ \psi\notin\{\phi_n\}\). Then, \(T_1(T_2\psi)=T_2(T_1\psi)=T_2(\lambda\psi)=\lambda(T_2\psi)\).

This mean \(T_2\psi\) is an eigenvector of eigenvalue \(\lambda\), so there is \(\mu\in\mathbb{C}\) such that \(T_2\psi=\mu\psi\).

Then, \(\psi\) is an eigenvector of \(T_2\) with eigenvalue \(\mu\).
\(\psi\) is a common eigenvector of \(T_1\) & \(T_2\), so \(\psi\in S\).
\(\psi\in S^\perp\), so this leads to \(\psi=0\), contradiction.

④ conclusion \(\therefore\) \(S=H\), and \(\{\phi_n\}\) is the orthonormal basis for \(H\) which is the eigenvectors of both \(T_1\) and \(T_2\).

(b) A linear operator \(T\) is normal if \(TT^*=T^*T\). Show that if \(T\) is normal and compact, then \(T\) can be diagonalized.

proof
①
\[\frac12(T+T^*)+i\frac{1}{2i}(T-T^*)=\frac12T+\frac12T^*+\frac12T-\frac12T^*=T\]

② \(\frac12(T+T^*),\ \frac{1}{2i}(T-T^*)\) commute

Let \(T_1=\frac12(T+T^*),\ T_2=\frac{1}{2i}(T-T^*)\). \[\begin{aligned} T_1T_2 &=\frac{1}{4i}(T+T^*)(T-T^*)\\ &=\frac{1}{4i}(TT-TT^*+T^*T-T^*T^*)\\ &=\frac{1}{4i}(TT-T^*T+TT^*-T^*T^*)\quad(\because\ T\ \text{is normal})\\ &=\frac{1}{4i}\big(T(T+T^*)-T^*(T+T^*)\big)\\ &=\frac{1}{4i}(T-T^*)(T+T^*)=\frac{1}{2i}(T-T^*)\frac12(T+T^*)=T_2T_1. \end{aligned}\]

③ \(T_1,\ T_2\) is compact

\(T\) is compact, so for given bounded sequence \(\{f_n\}\subset H\) there is a subsequence \(\{f_{n_k}\}\) such that \(\lim_{k\to\infty}\|Tf_{n_k}-g\|_H=0\) for some \(g\in H\).

\(T^*\) is compact as well, so there is a subsequence \(\{f_{n_{k_i}}\}\) such that \(\lim_{i\to\infty}\|T^*f_{n_{k_i}}-g^*\|_H=0\) for some \(g^*\in H\). \(\{f_{n_{k_i}}\}\subset\{f_{n_k}\}\), so \(\lim_{i\to\infty}\|Tf_{n_{k_i}}-g\|_H=0\).

Then, \(\|(T+T^*)f_{n_{k_i}}-(g+g^*)\|_H\le \|Tf_{n_{k_i}}-g\|_H+\|T^*f_{n_{k_i}}-g^*\|_H\).
So, \(\lim_{i\to\infty}\|(T+T^*)f_{n_{k_i}}-(g+g^*)\|_H=0\).

Thus, \(T+T^*\) is compact, as well as \(T-T^*\), and \(T_1,\ T_2\).

④ \(T_1,\ T_2\) is symmetric
\[\begin{aligned}T_1^*&=\left(\frac12(T+T^*)\right)^*=\frac12(T^*+T)=T_1,\\ T_2^*&=\left(\frac{1}{2i}(T-T^*)\right)^*=-\frac{1}{2i}(T^*-T)=\frac{1}{2i}(T-T^*)=T_2.\end{aligned}\]

⑤ conclusion \((\sim(a))\)

\(T,\ T^*\) is linear, so \(T_1\) and \(T_2\) are linear.

Then, using (a), there is an orthonormal basis \(\{e_n\}\subset H\) which are eigenvectors of \(T_1\) and \(T_2\): \(T_1e_n=\lambda_{1,n}e_n\) and \(T_2e_n=\lambda_{2,n}e_n\).

So, \(Te_n=(T_1+iT_2)e_n=(\lambda_{1,n}+i\lambda_{2,n})e_n=:\lambda_ne_n\).

\(\therefore\) \(T\) can be diagonalized by \(\{e_n\}_n\).

(c) \(U\) is unitary, \(T\) is compact, and \(U=\lambda I-T\). Show that \(U\) can be diagonalized.

proof
① \(T\) is normal

Using the property \(UU^*=U^*U=I\), \[\begin{aligned} TT^* &=(\lambda I-U)(\bar{\lambda}I-U^*) \\ &=|\lambda|^2I-\lambda U^*-\bar{\lambda}U+UU^*\\ &=|\lambda|^2I-\bar{\lambda}U-\lambda U^*+U^*U \\ &=(\bar{\lambda}I-U^*)(\lambda I-U) \\ &=T^*T. \end{aligned}\]

② conclusion

Due to (b), there is an orthonormal basis \(\{\phi_n\}\subset H\) s.t. \(T\phi_n=\lambda_n\phi_n\ \ \forall n\in\mathbb{N}\). Then, \(U\phi_n=\lambda\phi_n-T\phi_n=(\lambda-\lambda_n)\phi_n\ \ \forall n\in\mathbb{N}\).

Thus, \(U\) can be diagonalized with eigenvectors \(\{\phi_n\}\) and the corresponding eigenvalues \(\{\lambda-\lambda_n\}\).